area bounded by sinx, cosx, and x-axis

**yoman360** · April 19th 2010, 06:20 PM

find the area of the region whose boundaries are given

In the first quadrant, bounded below by the x-axis and above by the curves of y=sin(x) and y=cos(x).

(A) $2-\sqrt{2}$
(B) $2+\sqrt{2}$
(C) 2
(D) $\sqrt{2}$
(E) $2\sqrt{2}$

This is what i did.
$\int_0^{\pi/4} cosx-sinx dx$

solve integral i get: $-1+\sqrt{2}$
which is not one of the multiple choice options above. I think I am not understanding what the question is asking.

**drumist** · April 19th 2010, 06:27 PM

Take a look at the graph of cos x and sin x (attached below).

The region looks like it should be split into two portions, one from 0 to pi/4, and another from pi/4 to pi/2. The function defining the top of the region is different for each one.

Can you see what the integral should be?

Edit: Oh, I see what you did. You are looking at the wrong region. Look at the one that touches the x-axis.

**yoman360** · April 19th 2010, 07:23 PM

Originally Posted by drumist

Take a look at the graph of cos x and sin x (attached below).

The region looks like it should be split into two portions, one from 0 to pi/4, and another from pi/4 to pi/2. The function defining the top of the region is different for each one.

Can you see what the integral should be?

Edit: Oh, I see what you did. You are looking at the wrong region. Look at the one that touches the x-axis.

I'm still confused

**yoman360** · April 19th 2010, 07:59 PM

the question does not make sense because how can it be below the x-axis and in the 1st quadrant?

**drumist** · April 19th 2010, 09:14 PM

Originally Posted by yoman360

the question does not make sense because how can it be below the x-axis and in the 1st quadrant?

No, it's bounded below by the x-axis. In other words, the x-axis is what forms the bottom-most boundary of the region. So the region is ABOVE the x-axis.

Does this make sense?

I've filled in the region it is referring to with the color green in the attached image.

**yoman360** · April 19th 2010, 09:21 PM

Originally Posted by drumist

No, it's bounded below by the x-axis. In other words, the x-axis is what forms the bottom-most boundary of the region. So the region is ABOVE the x-axis.

Does this make sense?

yes. but don't know how to do it. I tried to seperate integrals one from 0 to $\pi/4$ and another from $\pi/4$ to $\pi/2$ (as you posted above) but i still get incorrect answer.

**drumist** · April 19th 2010, 09:47 PM

OK. The first half is bounded above by sin x, and the second half is bounded above by cos x.

$\int_0^{\pi / 4} \sin x \, dx + \int_{\pi / 4}^{\pi / 2} \cos x \, dx$

$= [-\cos x]_0^{\pi / 4} + [\sin x]_{\pi / 4}^{\pi / 2}$

$= -\cos (\pi/4) -(- \cos 0) + \sin (\pi / 2) - \sin(\pi / 4)$

You can then reduce this down to one of the answer choices that you listed.

Let me know if you can't follow what I am doing here.

**yoman360** · April 19th 2010, 09:54 PM

Originally Posted by drumist

OK. The first half is bounded above by sin x, and the second half is bounded above by cos x.

$\int_0^{\pi / 4} \sin x \, dx + \int_{\pi / 4}^{\pi / 2} \cos x \, dx$

$= [-\cos x]_0^{\pi / 4} + [\sin x]_{\pi / 4}^{\pi / 2}$

$= -\cos (\pi/4) -(- \cos 0) + \sin (\pi / 2) - \sin(\pi / 4)$

You can then reduce this down to one of the answer choices that you listed.

Let me know if you can't follow what I am doing here.

thanks

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